# Solving a system by using Cholesky Decomposition $(LDL^T)$

$$\begin{bmatrix} 4 & 1 & 1 \\ 1 & 2 & -1 \\ 1 & -1 & 3 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 3 \\ 1 \\ \frac{3}{2} \end{bmatrix}$$

Show that the matrix above is positive definite and solve the system using the decomposition $$A = LDL^T$$ with $$L$$ unitary angular and $$D$$ diagonal

I think I can use the decomposition below:

So we have

$$\begin{bmatrix} 4 & 1 & 1 \\ 1 & 2 & -1 \\ 1 & -1 & 3 \end{bmatrix} = \begin{bmatrix} D_1 \\ L_{21}D_1 & L_{21}^2D_1+D_2 \\ L_{31}D_1 & L_{31}L_{21}D_1+L_{32}D_2 & L_{31}^2D_1+L_{32}^2D_2 + D_3 \end{bmatrix}$$

But what do I do with the other coefficients of the matrix which are not equal to the right side? I mean, what do I do with $$a_{12}, a_{13}$$ for example?

And how do I proceed to solve the system?

If you have $$LDL^Tx=b$$, multiply on the left by $$L^T$$ to get $$DL^Tx=L^Tb.$$ So you have a system $$Dy=L^Tb$$, with $$D$$ diagonal; so you simply have $$y_j=(L^Tb)_j/D_{jj}$$ for each $$j$$. Since $$y=L^Tx$$, you have $$x=Ly$$, which you can calculate directly.
• Why multplying on the left by $L^T$ cancels $L$? – Guerlando OCs Apr 4 at 23:11
• Because $L$ is unitary. That's what you wrote? – Martin Argerami Apr 5 at 0:11